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Brilliant_brown [7]
4 years ago
15

Help lol

Chemistry
2 answers:
andreev551 [17]4 years ago
7 0
I would say high precision and not accuracy since even though they are close they are not all the exact number. Precision since they are close to each other, but not accuracy because they are not all the same exact number. (I would think, but I'm not totally sure.)
torisob [31]4 years ago
3 0
Maybe #3 both a high degree or precision and a high degree of accuracy;)
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3 years ago
g Calculate the pH when (a) 24.9 mL and (b) 25.1 mL of 0.100 M HNO3 have been added to 25.0 mL of 0.100 M KOH solution.
jarptica [38.1K]

Answer:

Following are the responses to the given choices:

Explanation:

For point a:

Using the acid and base which are strong so,

moles of H^+ (fromHNO_3)

= 24.9\ mL \times 0.100\ M \\\\= \frac{24.9}{1000\ L} \times 0.100\  M \\\\= 2.49 \times 10^{-3} \ mol

moles of OH^{-} (from KOH)

= 25.0\ mL \times 0.100\ M \\\\= \frac{25.0}{1000 \ L} \times 0.100 \ M \\\\\= 2.50 \times  10^{-3}\  mol  

1\ mol H^{+} \ neutralizes\  1\ mol\  of\  OH^{-}

So,  (2.50 \times 10^{-3} mol - 2.49 \times 10^{-3} mol) i.e. 1 \times 10^{-5} mol of OH^- in excess in total volume (24.9+25.0) \ mL = 49.9 \ mL i.e. concentration of OH^- = 2 \times 10^{-4}\ M

p[OH^{-}] = -\log [OH^{-}] = -\log [2 \times 10^{-4}\ mol] = 3.70

Since, pH + pOH = 14,

so,

\to pH = 14- pOH = 14- 3.70 = 10.30  

For point b:

moles of OH^- = from point a = 2.50 \times 10^{-3} \ mol

moles of H^+(fromHNO_3):

= 25.1 mL \times 0.100 M\\\\ = \frac{25.1}{1000}\ L \times 0.100 \ M\\\\ = 2.51\times 10^{-3} \ mol

1 mol H^+ neutralizes 1 mol of OH^-

So, (2.51 \times 10^{-3}\ mol - 2.50 \times 10^{-3}\ mol) i.e. 1 \times 10^{-5} \ mol \ of\  H^+ in excess in the total volume of (25.1+25.0) \ mL = 50.1\ mL i.e. concentration ofH^+ = 2 \times 10^{-4}\  M

Hence, pH = -\log [H^+] = -\log[2 \times 10^{-4}] = 3.70

6 0
3 years ago
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